Radically Compromised

It is given that:

$\begin{aligned} a_{n+1} & = \sqrt{2+a_n} \\ a_{n+1}^2 & = 2+a_n \\ \implies \frac {a_n}2 & = \frac {a_{n+1}^2}2 - 1 \end{aligned}$

We note that $\dfrac {a_{n+1}}2 = \sqrt{\dfrac {2+a_n}4}$ , as $\dfrac {a_1}2 < 1$ , then $\dfrac {a_n}2 < 1$ for all $n$ and we can let $\cos \theta = \dfrac {a_{n+1}}2$ .

$\begin{aligned} \cos \theta_n & = 2\cos^2 \theta_{n+1} - 1 \\ \implies \theta_n & = 2 \theta_{n+1} \\ \theta_{n+1} & = \frac {\theta_n}2 & \small \color{#3D99F6}{\text{As }\cos \theta_1 = \frac {a_1}2 = \frac {\sqrt 3}2 \implies \theta_1 = \frac \pi 6} \\ \theta_2 & = \frac {\theta_1}2 = \frac \pi {6\cdot 2} \\ \theta_3 & = \frac \pi {3\cdot 2^3} \\ \theta_4 & = \frac \pi {3\cdot 2^4} \\ \cdots & = \cdots \\ \implies \theta_n & = \frac \pi {3\cdot 2^n} \end{aligned}$

Therefore, we have:

$\begin{aligned} \log_2 \left(\cos^{-1} \left( \frac {a_{2016}}2\right) \right) & = \log_2 \left(\theta_{2016} \right) = \log_2 \left(\frac \pi{3\cdot 2^{2016}} \right) = \log_2 \pi - \log_2 3 - 2016 \end{aligned}$

$\implies a+b = 3+2016 = \boxed{2019}$